Modern Pratice of the Electrice Telegraph (Paperback)

This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1872 Excerpt: ...end. Multiply S by S and T by R, and add the products together; subtract from this amount T times S, and also R times S. Subtract the square root of the remainder from S; the remainder will give the resistance, x, or thedistance of the fault from the testing station. This process appears complicated, but is in reality very simple. For example, suppose the line 100 units lon, and the fault 68 units distant, and the resistance of tho fault 96 units, as shown in Fig. 53 X68 V32 We shall, however, have obtained these resistances by measurement, and not by calculation. We therefore have: S x S = 92 x 92 = 8464 ) 9. Rfl. T x R = 164 x 100 = 16400 p48b T x S = 164 x 92 = 15088 ) 24288 R x S = 100 x 92 = 9200 ) q And the square root of 576 is 24; which deducted, S = 92, gives 68 as the resistance of x, or the distance of the fault from the testing station. The distance, x, being known, the others are obtained with ease; for R--68 gives y, the distance from the opposite end; and T--68 gives z, or the resistance of the fault itself. This test should be taken from both See note, 104. ends of the line, if possible. In the above calculation the resistance of the fault is supposed to remain constant during the measurements; but as this is not often the case in practice, the average of several measurements should be taken. 129. To Find The Distance Of A Cross.--The two wires in contact form a loop, provided they are clean, and are twisted together, so that the contact offers no appreciable resistance. In such a case open both wires at the nearest station beyond, and test the resistance of the loop. Half this resistance will be the resistance of the wire between the galvanometer and the fault, and from this the distance can be calculated, as before explained (...